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i cant believe no one else has got the correct answer... people value of pi is 3.14
and @johrianshuman ur intersection is totally wrong
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Ans) xE [0,4]
x^2>=16
x E [-4,4]
sinx >=0... thus x is 0 to pi, 2pi tp 3pi, 4pi to 5pi....
x E [ 2 n pi , (2n+1) pi ]
intersection of the two is x E [0,pi]
*PLZ RATE* n joy u too
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units place of squares of all numbers end in one out of
0,1,4,9,6,5
for n to be a prime number , it cant have 0, 2, 4, 6, 8, 5 in units place ( except for n = 2 and 5)
thus its square cant end with 0, 4, 6, or 5 (except for n = 2.. as 2 is the only even prime number and n = 5 as mentioned in previous line)
thus we are left with 1 and 9 .. n square can only end with 1 and 9
and when we have this condition we see that for any such value of n we always have one composite number (non prime) out of 4n^2 + 1 and 6n^2 + 1
because with last digit as 1, 4n^2 + 1 will have 5 as units place
and with last digit as 9, 4n^2 + 1 will have 5 in units place
thus we are only left with n = 2 and n = 5.
but at n = 2 , 6n^2 + 1 becomes 25 which is not prime
thus the only solution is n = 5
*PLZ RATE*
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prime number is odd?!!!! who u kidding man
n = 2 is even n prime
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x/ y is generally the representation of a fraction ,,,
fraction raised to power decreases as the power increases
1/2 1/4 1/16 etc.
so limit x/y to power n n tending to infinity will be 0
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(4 - 9/2 )sq = ( 5 - 9/2 )sq
u dont solve this equation by jus takin sq root u have to bring terms to one side to form asquare - bsquare type of equation
and u get 5 - 9/2 = 9/2 - 4
which is true as 9/2 is 4.5 and it lies between 4 and 5
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root of ( (2root3) sq + 2 sq )
= root 12
or 2root3
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either alkene forms when elimination takes place CH3 CH = CH CH3
or when substitution SN2 occurs CH3CH-OH CH2CH3
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arre this is simple taara :)
if x = a is a root then (x-a) is a factor of the polynomial.. and since it is m times thus it is (x-a)^m is a factor of the polynomial
and thus F(X) (of degree n ) = (x-a)^m (g(x))
where g(x) is of power( n - m) because total degree of equation in left and right must be same which is n
*PLZ RATE* :)
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no... x^2 - y^2 = a^2 + b^2
by solving the two ( square both equations and add)
we finally get
(a^2 + b^2)sec^2 @ = (x^2 - y^2)sec^2 @ + 2y^2
thus for @to be eliminated the sec terms' coefficients must be same and thus we get the answer as
x^2 - y^2 = a^2 + b^2
*PLZ RATE*
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i think alkene forms CH3 CH = CH CH3
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yes CH3CH2OH undergoes mild oxidation during the process to form CH3CHO which has CH3C=O linkage and hence gives positive iodoform test
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see it this way
inf + 2 is also equal t o infinity
so infinity - infinity can be written as infinity - ( infinity + 2)
so it will be - 2 ... it can be any number this way.. got it ? :)
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neeral is correct and also infinity - infinity is not zero because we dont know the exact value of infinity... its just a concept of mathematics... infinty is a whole big branch of maths in itself its not just any simple number..
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